# How to prove $\frac{\partial z}{\partial \bar{z}} = 0$ if and only if $\frac{\partial \bar{z}}{\partial z} = 0$

Here is a problem form my complex analysis HW.

Unfortunately, I really have no idea how to go about this. Specifically, I don't really know how to take partials of that form. Does anyone have anything that might help me?

• The equivalence in the title is true because both statements are true. I refuse to discuss any problem containing the grotesque statement that $z$ and $\bar z$ are independent variables. – Georges Elencwajg Apr 15 '14 at 17:54
• @Georges: the relative notion of dependence here is almost surely "there exists a nontrivial differentiable function such that $f(z, \bar{z})$ is identically zero". – user14972 Apr 15 '14 at 17:59
• er, I meant "relevant" not "relative". And yes: $z$ and $\bar{z}$ are differentiably independent. They are clearly not functionally independent, but that's not the only sense that matters. Even in your example, we have examples of different kinds of dependence: $x$ and $x^2$ are linearly independent, for example. – user14972 Apr 15 '14 at 18:41
• @Hurkyl Though I feel like, unless otherwise stated, independent refers to functionally independent. – JLA Apr 15 '14 at 19:01
• @Georges: Wait a minute, I'm confused by the point of your example: your function relates $x$ and $x^2$, and thus proves them dependent in a variety of senses (e.g. functionally, algebraically, smoothly), not independent. – user14972 Apr 15 '14 at 19:04

Hint: If $z=x+iy$, usually you define: $$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$$ $$\frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$$ Can you take it from here?
You can compute $\mathrm{d}z$ and $\mathrm{d}\bar{z}$ in terms of the basis $\mathrm{d}x$ and $\mathrm{d}y$.
Then you can compute $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ in terms of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$.
Combining $\mathrm{d}z$ with $\frac{\partial}{\partial \bar{z}}$ gives $\frac{\partial z}{\partial \bar{z}}$ (using, for example, that combining $\mathrm{d}x$ with $\frac{\partial}{\partial x}$ gives $1$), and similarly for the other one.
I don't think the problem, or the suggested method, actually make sense. If $(z, \bar{z})$ truly to be interpreted as a "coordinate system", then the values of $\frac{\partial z}{\partial \bar{z}}$ and the other three combinations would follow from the usual conventions of what partial differentiation notation actually means (i.e. "take the derivative in the direction where this 'coordinate' increases and the other one is held constant"), and have nothing to do with whether or not $z$ is a "holomorphic coordinate".
• I don't get it either, $z$ and $\bar{z}$ are clearly not independent. – JLA Apr 15 '14 at 18:16
• @JLA: They are independent in many important senses; e.g. there is no complex differentiable function such that $\bar{z} = f(z)$, or even $f(z,\bar{z}) = 0$. The differentials $\mathrm{d}z$ and $\mathrm{d}\overline{z}$ are $\mathbf{C}$-linearly independent. And so forth. The goal of working with the variables $z$ and $\bar{z}$ is that they are independent in enough of the ways that actually matter, and calculus is much simpler when you can leverage complex differentiation. The formulation of your homework problem just doesn't make sense, at least not without being familiar with the book. – user14972 Apr 15 '14 at 18:41